/math/: Math general

if i pulled off the riemann hypothesis would you die

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The was posted recently, but it's an excellent problem. Solve for alpha only using classical geometry (no trig).

All right, here's some linear algebra to warm up:
Let [math]f \in \mathcal L(\mathbb C^n)[/math]. Prove that [math]f[/math] is diagonalizable iff [math]f^2[/math] is diagonalizable and [math]\ker f = \ker f^2[/math]. What can be said if instead [math]f \in \mathcal L(\mathbb R^n)[/math] ?

>>8258957
I would be extremely grateful

>>8258960
Don't all angles in a quadrilateral add to 360?
Just add the angles up lmayo

>>8259332
There are two unknowns, not just one.

>>8258960
b=1
a=129

prove me wrong

>>8258960
apply the inscribed quadrilateral circle theorem
and you can solve the rest

>>8259387
which means

a=100-30
a=70
Therefore 50+70+b=180
b= 60

>>8259203
We suppose that [math]\operatorname ker f = \operatorname ker f^2[/math] and [math]f^2[/math] is diagonalizable. The assumption on kernels allows us to assume WLOG that 0 is not an eigenvalue of [math]f[/math] through a standard calculation with Jordan form. Suppose we have a [math]v \in V, k>0, \lambda \in \mathbb C^*[/math] for which [math](f - \lambda I)^k v = 0[/math]. Observe that [math](f^2 - \lambda^2 I)^k v = \left( (f + \lambda I) (f - \lambda I) \right)^k v = 0[/math] using that these two operators commute. Thus [math](f + \lambda I) (f - \lambda I) v = 0[/math] because [math]f^2[/math] is diagonalizable, so all eigenvalues are in fact proper. We conclude that [math](f - \lambda I) v = 0[/math] (I don't have time to expand on this), so [math]f[/math] is indeed diagonalizable.

if i have a cube that is 100m^3
how many smaller cubes could fit inside it if they werent?

>>8258951
Newfag fucker here.

I want to learn mathematics out of sheer curiosity. How deep does the rabbit hole go? Is it worth the effort?

How do I approach my math major optimized for graduate school towards getting a job and a career outside of academia? How do I start graduate school and leave with an industry job?

>>8259518
yes

>>8259387
Scrub here. Mind explaining that briefly?

Let [math]f:\mathbb{C}\left[ {x,y} \right] \to \mathbb{C}\left[ {x,y} \right][/math] be the map such that [math]f\left( {{x_i}} \right) = {f_i}[/math] and [math]\det \left( {\partial {f_j}/\partial {x_i}} \right) \in \mathbb{C} - \left\{ 0 \right\}[/math].

Show [math]f[/math] is an isomorphism.

>>8259471
Nice, that'll work.

>>8259580
As far as you are willing go. Obviously, no one has seen the end

Some analysis:
Characterize the functions [math]f: \mathbb R \to \mathbb R[/math] such that, for each [math]x[/math] and each sequence [math](u_n)[/math] such that [math]\left(\frac{u_1 + \dots + u_n}{n}\right)[/math] converges to [math]x[/math], [math]\left(\frac{f(u_1) + \dots + f(u_n)}{n}\right)[/math] converges to [math]f(x)[/math]

>>8259583
Publish papers which have applications to industry; theoretical comp sci oriented and the like, and you should recieve job offers towards the latter end of your phd. If youre getting a masters, just apply for interviews; math degrees are highly employable.

>>8259614
An isomorphism of vector spaces ? of rings ?

>>8259518
No, because if the cubes want water

Trying to understand cyclical linear logic (CyLL). If it's modeled by a (nonsymmetric) monoidal category, why does the cyclic rule not imply exchange? You have [math]A \otimes B \cong B \otimes A[/math] so if you have associativity then you have [eqn](A \otimes B) \otimes C \cong (B \otimes A) \otimes C \cong B \otimes (A \otimes C) \cong B \otimes (C \otimes A) \cong (B \otimes C) \otimes A \cong (C \otimes B) \otimes A[/eqn] which isn't a cyclic permutation. Or is it modeled differently?

>>8259699
under the assumption that multiplication is commutative

>>8259717
Huh?

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For random variables I am trying to understand what the below equation means:

Pr[X=v] := Pr[X^-1(v)]

Is it basically saying the probability of v in set V is the same as the probability of falling into the pre-image of v in set U?

Here's a problem i've been thinking about:
Let's consider strings of digits, which all have the same length n. There are 10^n such strings.
From those 10^n strings we pick a set of strings S, such that any two strings in S:
(a) differ at least on two places
(b) differ at least on three places
(c) differ at least on k places
What is the maximum possible number of elements in S? I can prove that for (a) it's 10^{n-1}, but i don't know about the others.

>>8259717
Basically my question is how do you model the cyclic permutation of two things without making the product commutative? Even if you do something like [math]Hom(A, B \otimes C) \cong Hom(A, C \otimes B)[/math] that should imply commutativity by Yoneda.

>>8258951
Is it possible to use only chords to divide a circle into equal area pieces with no two pieces congruent?

>>8259203
Why make the distinction with linear maps over real vectorspaces?
Or does diagonalisable mean real eigenvalues there?

>>8259767
Of course. Do you mean "with ruler and compass only"?

>>8259771
Yes, it means diagonalizable with real eigenvalues

>>8259203
I see a solution was posted already but I just want someone to check my reasoning:
[math]f[/math] is diagonalizable iff its jordan matrix is diagonal.
If f is diagonalizable, then the implication holds easily.
If f is not diagonalizable, then it has a 2x2 (or bigger) jordan block. If this block has nonzero eigenvalue, then the jordan matrix [math]f^2[/math] contains exactly this block squared, so [math]f^2[/math] is not diagonalizable. On the other hand, if the block had eigenvalue = 0, then kernel of [math]f^2[/math] is bigger than ker f.

>>8259636
These are exactly the linear functions. It's easy to check that these work.
Now assume that f is not linear; then we can find three points [math]a<c<b[/math], such that [math]c = ta + (1-t)b[/math], but [math]f(c) \neq tf(a) + (1-t)f(b)[/math].
Then create a sequence containing [math]a[/math] and [math]b[/math] in appropriate proportion. It will fail to converge to [math]f(c)[/math].

>>8259330
You're a smart guy

I wish we had an irc or skype group or something so we could do higher level textbooks together as a group-study/group-mentoring community thing. I like learning with other people. I wanted to do some stuff like sheaves and schemes or something. Of course open to all topics at the 4th year undergrad/grad level.

>>8259811
I'm down, do you wanna create it?

>>8259828
Sure, skype or irc?

>>8259829
Up to you. Maybe both if you want, then link to each other in the info.

I'd help but I'm at work right now and I don't get off for another 8 hours.

>>8259840
>>8259829
>>8259828
I can create a irc channel and post the info here later. I will be able to do it after I finish studying. It won't be for another few hours.

>>8259849
Scratch that, I am making one now. Will post info soon.

>>8259854
Sounds good, I will see if I can get irc working on my phone.

>>8259854
If you are truly busy I can make one.

>>8259863
>>8259868
I'm part of a group that's already been doing this (as of around the middle of March), would you like an invite? Just a slight issue though, we're not using IRC.

>>8259877
Oh, whatcha usin?

>>8259877
let's use irc for now, it's simple to use and anyone can join.

>>8259878
We're using a slack group consisting of multiple channels dedicated to projects/subjects of study.

>>8259878
Unfortunately we're using Slack. It's sort of like a glorified IRC but it helps us have nice features like private channels and etc.

>>8259882
Does that cost money?

>>8259723

That seems right. The probability of outcome v for random variable X is the same as the sum of probabilities over all arrows leading to v.

The function X is not injective and not necessarily even surjective (all points could map to 1, i.e. X is always true), while X^(-1) should be both surjective (every point in U has an outcome) and injective (each point in U has only one assigned outcome).

Please check this yourself, don't take my word for it.

>>8259890
No it's free, there are paid options but the one we're using is completely free.

>>8259890
No, it's a free service. If you're interested in joining, send us an e-mail. (I recommend you create a throwaway email because I wouldn't post my personal email on here)

>>8259780
is it? show please

>>8259893
Thanks!

>>8259901
Draw one chord followed by a perpendicular bisector of that chord using arcs and a straight-line or something followed by yet another bisector of that bisector. This is my best guess.

>>8259905
you guys should come to the slack the others are commenting
it's a nice interface and simple enough to use
we've had people going through math for a while

>>8259906
I don't think that's going to work

>>8259908
Oh shit I didn't notice that anon said "no conrguent pieces".

>>8259907
I heard it was only basic algebra and logic with only a few people doing functional.

If [math]A[/math] is involutory, show [math]\frac{1}{2}(I+A)[/math] and [math]\frac{1}{2}(I-A)[/math] are idempotent, and [math]\frac{1}{2}(I+A) \frac{1}{2}(I-A) = 0[/math].

You should be able to solve this.

>>8259797
The square of a matrix in Jordan form need not be in Jordan form, I believe. Consider

[math]\displaystyle \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} [/math]

>>8259916
it's hard to say what "a few people" means, because it's a small group.

it's divided in channels though, so if you stick to the math channel you only talk to math people, simple.

i say it's fine to have a separate slack and irc channel.

so far everyone in the irc channel is doing higher level math and no plebs

>>8259912
with congruent pieces in the sense of circular sectors it won't work anyways
for example you can't construct a regular heptagon with only circle and straightedge, which is equivalent to finding 7 equal circular sectors

>>8259916
That doesn't matter much anyway, since the reason why the discussions are rather low-level is because most people are exactly at that level. The slack group is just a way for people to gather and study together. If grad students come together, then they'll talk about grad mathematics. We can always create new projects if people are interested.

>>8259929
So how many people are working on functional then? Exactly?

>>8259940
two I think, maybe someone else is

>>8259941
Yep around that out of the five-six people that are in galois (the higher-level math channel). The last project was, well, galois theory.

[eqn]P = \frac{v^{\top} v}{vv^{\top}}[/eqn] Prove [math]P^2 = P[/math].

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>>8259901
You might try this sort of thing. I'm not sure is there any solution but why not))

>>8259951
v * vtranspose is symmetric

so is vtranspose * v

symmetric means equal to transpose

(vtranspose*v)transpose = v*vtranspose

because v*vtranspose and vtranspose*v are symmetric and tranpositions of each other they are the same.

P = I, I^2 = I

>>8259982
[math]v[/math] is a vector, not a matrix. The result is a scalar.

>>8259951
I guess you mixed up the transposes.
There is a shorter way to prove it, but I like this one, since it shows, what kind of map [math] P [/math] actually is:

Take any orthogonal base of vectors including [math] v [/math].
For any vector we have
[math]Pw= \frac{vv^T}{v^Tv}w = \frac{\left \langle v,w \right \rangle}{v^Tv} v [/math]
and with that we have [math] Pv = v [/math] and [math] Pw = 0 [/math] for every other base vector.
This also gives us [math] P^2v = P(Pv)=P(v) [/math] and [math] P^2w = P(Pw)=P(0) = 0 = Pw [/math]

>>8259981
Hm, sorry, that one doesn't work.

>>8259614
That is the Jacobian Conjecture...

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>>8260037
>>8259901
This might work

>>8259801
for you

>>8260103
okay, you might have done it for 7.
now show a general straightedge and compass construction for n pieces

>>8260156
freenode's is a aimed for math sci's

Is anyone into HoTT? Your thoughts? Is it another fancy idea or does it have immediate applications allowing one to easily prove easy things?

>>8258957
It would be extremely painful (like when Wiles pulled off Fermat's last theorem).

>>8258960
there is no alpha

>>8260508
hes a smart guy

[email protected] for the slack group please

>>8260205
bumb

Why is {(x, y) | x is an odd integer, and y is an even integer} not a function?

>>8261207
a function has one output for every input
where would 1 go under that rule? that set has (1,2) (1,4) for example, so multiple outputs

>>8261217
I see...I hadn't thought of a function that would output two different values I was thinking in terms of one output...
Thnx

>>8261224
That's the thing. The entire notion of a function captures the idea of assigning a unique output to every input. So given some guy in the domain, I can speak unambiguously about its value in the image.

>>8259921
Just write down the definitions bro.

[math]\frac{1}{2}(I + A)\frac{1}{2}(I + A) = \frac{1}{4}(I + A + I + A) = \frac{2}(I + A)[/math] etc.

>>8261550
oops, that last term should be [math]\frac{1}{2}(I + A)[/math]

Combination formula:

The calcul for find k-combination of a set S who has n elements is: n! / k!(n-k)!

Whether a set with 3 elements (a, b, c): there are 6 combinations of 2 elements: ab, ac, ba, bc, ca, cb.

But with the formula: 3!/ 2!*1! = 6/2 = 3, so 3 elements.

What is wrong ?

>>8261556
>>8261556
>Whether a set with 3 elements (a, b, c): there are 6 combinations of 2 elements: ab, ac, ba, bc, ca, cb.
Those are permutations. Combinations are where order doesn't matter, so ac = ca.

Consider the xy plane and color every integer coordinates with 1 of 3 colors. Prove that there exists a rectangle such that all 4 corners are the same color.

>>8260205
No one cares about true rigorous computer math? :(

>>8261605
Right now it mainly has practical applications for some mathematicians, like people who want to prove things about infinity groupoids and homotopy. They've found new proofs of old results like [math]\pi(S^1) = \mathbb{Z}[/math] which is pretty cool.

Then again, if Mochi had proved abc in Coq, no one would have been able to dispute its correctness. It decouples proof verification from understanding (which obviously has its pros and cons). But of course you don't need HoTT for that.

>>8261602
Consider only a finite subset of the plane: points [math](x, y)[/math] with [math]x, y \in \mathbb{Z}[/math] and [math]0 \leq x \leq 3^4, 0 \leq y \leq 3[/math].
So you have vertical columns, each consisting of 4 points. There are [math]3^4 [/math] ways to color a column with three colors, but we have [math]3^4 + 1[/math] columns. So you can pick two columns colored the same way. From 4 points in this column, you can pick two points with the same color. These two points, together with corresponding points ono the other column make a rectangle.

someone posted this like a year ago and i thought it was pretty cool:

Find the solutions of
[math] \frac{1}{a^2}+ \frac{1}{b^2}= \frac{1}{c^2} [/math] for [math]a,b,c \in \mathbb{N} [/math]

>>8260949
UUUU

>>8261657
cant find any desu

>>8261994
hint:
Try to find conditions such that c is an integer.

>>8261657
First we can solve equation
[eqn] \frac{1}{a} + \frac{1}{b} = \frac{1}{c} [/eqn]
All solutions are of the form [math]a = mx(x+y), b = my(x+y), c = mxy[/math] for some natural numbers m, x, y, gcd(x,y) = 1
Now we want all these three numbers to be squares. By dividing things, we see that [math]\frac{x}{y}[/math] and [math]\frac{x+y}{y}[/math] are squares of rationals. Also gcd(x, y) = 1 so we have [math]x = x_1^2, y = y_1^2, x+y = z_1^2 [/math].
So simply all triples (x, y, x+y) are pythagorean triples, so by the well known parametrization [math]x = s^2 - t^2, y = 2st[/math].
Going back to the top we get the final formulas.

f.e. (16, 9, 12) is a solution.

>>8262065
1/a^2+1/b^2=1/c^2
so c=ab/sqrt(a^2+b^2)
now we know for integral a and b and if sqrt(a^2+b^2) is element of N, that we can substitute a=2t, b=(1-t^2)
so we search t such that 2*t*(1-t^2)/(1+t^2) is element of Z
but I complied in a small code and didnt find any solution smaller than 50000
>>8262088
16,9,12 is not a solution ?

>>8262110
Sorry, i wrote down the wrong numbers.
The general formula is
[math]a = s^4 - t^4, b = 2st(s^2 - t^2), c = 2st(s^2 - t^2) [/math] (or all multiplied by a constant)
Plugging in s=2, t=1 gives triple (15, 20, 12) which is a solution.

i have a question: >>8262185

[email protected] for the slack group
I think I have been in this group but i think I have been inactive for too long (busy year)

>>8258951
Today I spent an entire hour in class proving a couple of theorems using Peano arithmethic to have a rigorous basis with which I could use to then solve equations involving natural numbers of the form

a + b = c

While the engys were doing integrals and the physicists were doing analytical geometry I was proving that x + 2 = 15 if and only if x = 13.

I feel so good right now, this major is so fucking ridiculous (in a good way) I literally can't even.

>>8259797
Second proof is clever, I like it (the one I had in mind was a bit longer). I did not understand the jordan block argument in the first proof though.

Another analysis exercise:
Let [math]f[:[0,+\infty[ \to \mathbb R[/math] a smooth (ie. infinitely differentiable) function such that [math]\displaystyle f(x) \underset{x \to +\infty}{\longrightarrow} f(0)[/math].
Prove that, for each [math]n \ge 1[/math], [math]f^{(n)}[/math] vanishes.

>>8262199
i could still need help / opinions

>>8259780
No, any chords at all. Let's see you try.

>inb4 doesn't know what chord means

>>8260103
It almost works but it does not distribute areas evenly. Good job though on finding the simplest candidate which breaks symmetry.

>>8262268
vanishes where

>>8262446
well somewhere, but it doesn't have to be at the same place for each derivative

>>8262110
>>8262130
There's an easier way to find solutions.

For every Pythagorean triple A^2 + B^2 = C^2, there exists a solution a, b, c where
a = BC
b = AC
c = AB.

An example would be a = 20, b = 15, c = 12. 156, 65, 60 is another solution. 255, 136, 120 is another.

>>8262571
so [math] \forall n \geq 0 ~ \exists x : f^{(n)}(x) = 0 [/math]?

>>8262710
[math] n \geq 1 [/math] of course

>>8259370
so use the right side triangle and make another equation adding up to 180.

2 linear equations in 2 variables, snore

>>8262088
>>8262110
>>8262130
good job
my solution back than was basically like >>8262617
but it misses the non existence of other solutions which was done by >>8262110 even though he fucked up the parametrization of pythagorean triplets, which is why his code didn't work
here's how you generate pythagorean triplets
https://en.wikipedia.org/wiki/Pythagorean_triple#Points_on_a_unit_circle

>>8262730
Well in return he missed a general formula to find all solutions so I guess having both components are good.

I didn't bother with finding the nonexistence of other solutions because he already did it.

>>8262217

I truly hope this is satire.

I flip coins until I get 3 heads in a row. What's my expected number of coinflips?

>>8262836
2

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>>8262217

>>8262863
Don't think so.

>>8262836
it's 2

>>8262922
It's going to be at least 3 since you need to flip at least 3 times to get 3 heads in a row.

>>8262951
It's 2 though. Flip twice. If both result in heads, keep it heads. Now you have a total of three heads.

>>8259332
>>8262714
These posts are /sci/ in a nutshell. Condescendingly suggest an idea that doesn't work, see someone point out why it's wrong, update it to a slightly different idea that still doesn't work, and then dismiss the problem as trivial and boring without even realizing you've failed to solve it. Well done.

>>8262710
Exactly

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>>8262999
But that makes no sense

>>8262877
>>8262782
what's wrong ? I'm not really interested in "foundations of mathematics", but I can totally see how someone can find it fascinating

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>>8258951
Hi, today I've tried to compute areas using gauss-bonnet formula. It's easy to compute sphere area since gauss-bonnet becomes Area/r^2=4pi.

But then I decided to find the area of a spherical cap and got a problem because there could be two caps with different area but identical boundary (pic related). How can these boundaries have different geodesic curvature?? They obviously equally deviate from being geodesic, well because they are equal.

>>8263238
Hmm I can kind of see it working for the first derivative, since we can make an argument along the lines of the mean value theorem (needs a little fine tuning since we're not on an interval)
For higher order derivatives it seems to get really icky, since it won't even hold on any intervalls (take for example f as the solution of the poisson equation with nonzero right hand side)
Any hints?

>>8263282
Well try to make a good case for why it works for n=1, and more importantly n=2 and hopefully you should have a better idea of what the general proof could look like. Otherwise I'll give you more info

>>8263325
For [math]n=1[/math] I would take
[eqn]
\int _0^\infty f'(x) dx = \lim_{x\to \infty}f(x) -f(0) = 0
[/eqn]
If [math] f'(x) > 0 [/math] for some [math]x[/math], then by the continuity of [math] f' [/math] there's a whole open region around [math]x[/math], where [math]f'[/math] is positive,
which would result in a positive integral, if there wasn't also a region, where [math]f'[/math] is negative.
The intermediate value theorem then gives us existence of an [math]x[/math], s.t. [math]f'(x) = 0[/math]

Unfortunately I don't see this kind of argument working out for [math]n>1[/math]

>>8261994
You can actually apply Hilbert's Theorem 90 to find an exact form that all solutions must satisfy.

>>8263437
Well let's scratch our heads then. Assume that [math]f''[/math] does not vanish (ie. basically f is convex), what can the graph of f look like, knowing that f' vanishes ? Is that compatible with the assumption that f converges to f(0) ?

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I know you guys hate posts like this but can anyone help me with the 2nd question . my dad is gonna beat the shit out of me if i dont figure it out

>>8263851
>dad is gonna beat the shit out of me if i dont figure it out
Well maybe he should, because this is really not that hard. Just expand the factorial and see that it works

>>8263824
fucking hell - just deleted my whole post.
Appearently it's not a good idea to write everything in the tex preview window.

So once again:
If [math] f'' [/math] doesn't vanish, we can asume wlog that [math] f [/math] is strictly convex.
Since there exists a [math] x_0 [/math], s.t. [math] f'(x_0) = 0,~f [/math] has a minimum there.
we also know that for every [math] x_1 > x_0 [/math] we will have [math] f'(x_1) >0 [/math].
Together with the fact, that the graph of [math] f [/math] must always lie over the tangent at [math] x_1 [/math] due to [math] f [/math]'s convexity, we have
[eqn]
f(x_1+t) > f(x_1) + f'(x_1)t
[/eqn]
for all [math] t>0 [/math] and thus [math] \lim_{x\to\infty}f(x) = \lim_{t\to\infty}f(x_1+t) =\infty [/math] which is a contradiction.

Now if we wanted to make a similar argument for [math]f^{(n)} [/math], we would have to show first,
that [math] \lim_{x\to\infty} f^{(n-2)}(x) \neq \infty [/math], which I don't see a reason for to be honest

How do I convert from barycentric coordinates in an n-simplex to n-dimensional Cartesian coordinates? I'd really appreciate an answer in a form that could be easily incorporated into some code I'm writing.

>>8264019
For this definition it seems like you can just sum up the baricentric coordinate coefficients multiplied with their respective simplex corner vectors and get your vector in cartesian coordinates.
In matlab for example I would write this as a matrix multiplication:

Take barycentric coordinate vector a.
Create matrix X that contains the vectors of your simplex corners x_i in cartesian coordinates.
Multiply X with a and you'll get the vector in cartesian coordinates

>>8264015
All right, great. Now for the general case, just modify this slightly to prove by contradiction that if [math]f^{(n)}[/math] vanishes, then [math]f^{(n+1)}[/math] also vanishes, which will prove the desired result by induction.

>>8264042
But I thought that barycentric coordinates were in terms of enclosed area/ volume/ nD space so they mapped nonlinearly onto Cartesian coordinates. Is it really that simple? Am I an idiot?

>>8264077
The barycenter is just the point for which the distance to each vertex is 1. It has nothing to do with volume. Why you think it does is probably because the name means the center of weight, and thus you think of physics and so eventually volume.

If you want to convert the coordinates, I'll give you a tip! Each n-simplex is homeomorphic to the so called "standard" n-simplex, the one in the n-dimensional Euclidian space. It's vertices are (1, ..., 0), (0, 1, ..., 0), ..., (0, ..., 0, 1). This should get you started.

Are preliminary exams within a math course common at the graduate level? I'm starting my Master's later this month and while I'm taking Stats, we're having a prelim exam the first day. I'm a bit worried since I haven't taken a Stats course since Spring 2015 (so I don't remember much of anything), and the entire concept itself kind of rubs me the wrong way. I guess I'm mostly venting here.

>>8262268
I was wrong with the linear algebra solution - as someone pointed out, a square of jordan block is not necessarily a jordan block. I think the following is true instead: if f has a jordan block with nonzero eigenvalue x, then f^2 has a jordan block with the same size and eigenvalue x^2.

As for the analysis problem:
Imagine [math]f^{(n)}[/math] doesn't vanish. Then it's either positive everywhere or negative everywhere, otherwise we'd use intermediate value theorem.
So [math]f^{(n-1)}[/math] is strictly monotonic. Let's say it's increasing, if it's decreasing the same argument works.
Now there is a possibility that [math]f^{(n-1)}[/math] becomes positive at some point a. But then [math]f^{(n-1)}(x) > f^{(n-1)}(a) > 0[/math] for all x>a. From this we have contradiction, since [math]f[/math] would be going to infinity as x->infinity.
So we conclude that [math]f^{(n-1)}[/math] is negative everywhere (or positive everywhere in the other case)
But then [math]f^{(n-2)}[/math] is monotonic, and this way we go up to f. But f can't be monotonic (unless it's constant but then it's all trivial).

>>8264277
It's just to show yourself whether you are adequately prepared, since people will come in with vastly different backgrounds.

>>8264513
I figured so. I'm hoping it doesn't count for a fucking grade. I'm kind of doubting it will be but who knows. I know I'm not prepared but I was going to study the material during the semester and all that anyway.

>>8263276
In case someone is still interested in this, it's because the lines do not, in fact, have the same geodesic curvature. Imagine you walk in the lines in the same direction. One curves to the 'left' and one to your 'right'. They lines like a mirror imagine of one another, they are not the same tho.

>>8258951
What is the resistance between two resistors on a matrix of infinite variable resistors, whose variance is set by a mandelbrot set (fractal variance based on distance from primary), where the two resistors are seperated by three resistors length from the primary each.

>>8265238
They are all one ohm, btw....

>>8265241
>They are all one ohm, btw....

And powered by the relative motion of a 0.005 tesla magnetic field rotating at one rpm, at a distance from primary of 1m

>>8265245
Also, distance between resistor nodes, 1cm... coppoer contacts, resistor length .2cm and non electromagnetically conductive.

>>8264279
That'll work, yes

Here's some more:
Characterize the polynomials [math]P \in \mathbb C[X][/math] such that
1. [math]P(\mathbb U) \subset \mathbb U[/math], where [math]\mathbb U[/math] is the unit circle
2. [math]P(\mathbb Q) \subset \mathbb Q[/math]
3. [math]P(\mathbb Z) \subset \mathbb Z[/math]

Compute [math]\displaystyle \int_0^1 \frac{t-1}{\log t}[/math]

Let [math]\alpha \in [0,\pi][/math] and [math]u_n(\alpha)[/math] the number of changes of signs in the sequence [math]1, \cos \alpha, \dots, \cos(n\alpha)[/math]. Prove that [math]\displaystyle \frac{u_n(\alpha)}{n}\underset{n \to \infty}{\longrightarrow} \frac{\alpha}{\pi}[/math]

Access to local files on my machine is disabled by my administrator but basically its the Monty Hall Problem.

You have 3 Doors

/___\.../___\.../___\
|_?_|...|_?_|...|_?_|
|___|...|___|...|___|

Behind one of these doors is a Brand New Car, but behind the other two doors are only lobotomised sex slaves. Obviously you want the car, not the retards. You are about to select a door when one of the doors opens, and reveals a sex slave. Should you change your selection now that you know one of the doors is definitely not the car?

/___\.../___\.../___\
|_:<.|...|_?_|...|_?_|
|___|...|___|...|___|
............../\........./\...?

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>>8258951
So I wrote this little module in Haskell that calculates the integral of a function from point a to point b.

It works on every polynomial that I've tried it with, and also on the exponential function.

However, when I try to integrate sin(x) from 0 to pi I get 2 as the answer.

Does anyone know what the problem is here?

Also if you have other, better numerical methods for approximating integrals please share them.

>>8265676
Doing the last one:

The sequence is basically just the function [math] \cos(\alpha x) [/math] evaluated in the natural numbers. The number of changes of signs in one period is 2.
The number of changes of signs when we go from one to n will be [math] u_n= \left \lfloor 2n\frac{\alpha}{2\pi} \right \rfloor [/math]
(I'm actually not that sure if it should be floor or ceil or maybe round, but something along those lines, which will be unimportant in the limit process)
Therefore [math] \lim_{n\to\infty}\frac{u_n(\alpha)}{n} = \frac{\alpha}{\pi} [/math]

>>8265789
This seems pretty weird. It seems like there has to be a scaling error at some point, but that would also affect the integration of the polynomials.
In general a numerical integration scheme's quality is determined by how well it can integrate polynomials.
Unfortunately I don't have any experience with haskell, so I can't check your code.

Does anyone understand the matrix groups as lie groups? I was wondering if anyone could walk me through showing how SU(2) is diffeomorphic to a 3 sphere, and hence a lie group and how to find its lie algebra?

I really don't get it and my supervisor is not impressed.

>>8265789
>>8265806
oh well, wait a minute. the integral of sin from 0 to pi IS 2.
kek

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>>8265806
>>8265809
What the hell, why did I think it was 1?

Kek, thanks anon.

>>8265676
let me try, although I feel it will be wrong
1. f(x)=c*x^n and ||c||=1
2. f element of Q[x]
3. f element of Z[x]

>>8259760
>(b) differ at least on three places
This is not possible for n=2 right? AMD for n=3 the most elements in the string is 10. So I'm just gonna guess that the answer is 10^(n-2) and for (c) it's gonna be 10^(n-(k-1)) but I have no idea of how to formally prove this. Maybe start by saying for any n you can't have two strings that differ for more than n. Then go on to prove that the most possible elements in S will occur only if k=n. Then go on to prove that for k=n the most possible elements is 10.

>>8265825
update for 3
f from Q[x]
f(x)=1/k*g(x) ,such that g from Z[x],k from Q and g(Z/kZ)=0
(you basically divide the ggc out of f(x) )

>>8265807
Well one way to look at this is to note that the space of matrices [math]\mathbb H = \left{\begin{pmatrix} a & - \bar b \\ b & \bar a \end{pmatrix}, a,b \in \mathbb C \right}[/math] is a 4-dimensional real algebra and that the determinant defines an euclidean norm.
The unit sphere is exactly SU(2).
The Lie algebra is the tangent space of SU(2) at the identity matrix.
Let's compute that tangent space and be done with it: SU(2) is the set of matrices of [math]\mathbb H[/math] such that [math]\det A = 1[/math], which we also note is the set of matrices of [math]\mathbb H[/math] such that [math]\frac{1}{2}\mathrm{tr}(A* A) = 1[/math]. Now the tangent space of SU(2) at the identity is simply the orthogonal of the identity (that's true of every sphere), ie. in our case, the space of matrices of [math]\mathbb H[/math] with trace 0, ie. [math] \left{\begin{pmatrix} a & - \bar b \\ b & \bar a \end{pmatrix} | Re(a) = 0 \right}[/math]

>>8265890
[eqn]\mathbbH=\left\{ \begin{pmatrix}a&-\barb\\b&\bara\end{pmatrix},a,b\in\mathbbC\right\}[/eqn]

[eqn]\left\{ \begin{pmatrix} a & - \bar b \\ b & \bar a \end{pmatrix} | Re(a) = 0 \right\} \left\{ \begin{pmatrix} a & - \bar b \\ b & \bar a \end{pmatrix} | Re(a) = 0 \right\}[/eqn]

No need to thank me.

Protip: To show curly brackets in LaTeX, use \{ or \lbrace and \} or \rbrace.

>>8265894
Oops. Looks like there were more mistakes than I initially thought.

Or maybe LelTeX is acting up again. Who knows. Maybe it's both.

[eqn]\mathbb{H}=\left\{ \begin{pmatrix}a&-\bar{b}\\b&\bar{a}\end{pmatrix},a,b\in\mathbb{C}\right\} \\
\left\{ \begin{pmatrix} a & - \bar{b} \\ b & \bar{a} \end{pmatrix} | Re(a) = 0 \right\} \left\{ \begin{pmatrix} a & - \bar{b} \\ b & \bar{a} \end{pmatrix} | Re(a) = 0 \right\}[/eqn]

>>8265894
you seriously still haven't learned to use the tex preview button?

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>>8265905
What is shown during TeX preview and what LelTeX actually shows when you post it is sometimes different.

[eqn]\mathbb H = \left\{ \begin{pmatrix} a & - \bar b \\ b & \bar a \end{pmatrix}, a,b \in \mathbb C \right\}[/eqn]

[eqn]\left\{ \begin{pmatrix} a & - \bar b \\ b & \bar a \end{pmatrix} | Re(a) = 0 \right\}[/eqn]

See the difference?

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>>8265909
>now then it works
Well fuck you too.

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>>8263024
I hoped not. Back to /g/ then..

>>8258960
>classical geometry
help pls

13 mod -3 = 1
13 mod -3 = -2

Both are correct ?

>>8266123
[math] 1 \equiv-2 \pmod 3[/math]

>>8265676
1) Take any [math]z[/math] from the unit circle. Then we know that [math]|P(z)| = 1[/math]. But on unit circle [math]|P(z)|^2 - 1 = P(z) \bar{P(z)} - 1[/math] is a rational function of [math]z[/math], because [math]\bar{z} = 1/z[/math]. So we have a rational function in z which is zero everywhere on unit circle. So the function itself must be zero. Now assume that P(z) has more than one term - then it has the highest order term [math]az^n[/math] and the next term with nonzero coefficient [math]bz^k[/math]. Then it's easy to see that [math]z^{n-k}[/math] can't vanish from [math]P(z) \bar{P(z)} - 1[/math], contradiction. So the answer is [math]cz^n, |c|=1[/math].

2) Let's say the polynomial P has degree n. Then I look at n+1 any rational arguments, for example [math]0, 1, ... n [/math] and use Lagrange interpolation - from the formula it's obvious that P has rational coefficients. So the answer is simply polynomials with rational coeffs.

3) These are polynomials which can be written as
[eqn] P(x) = a_0 \binom{x}{0} + a_1 \binom{x}{1} + ... + a_n \binom{x}{n}. [/eqn]
It's easy to see that these work. To prove that these are all polynomials: let's say P has degree n, then it is easy to write down a polynomial P_1 that looks like above and has the same values as P for arguments 0, 1, 2...n. But then they agree on n+1 places so they must be the same.

>>8266168

Sorry, i don't understand your answer but i've find an answer on wiki:

"However, this still leaves a sign ambiguity if the remainder is nonzero: two possible choices for the remainder occur, one negative and the other positive, and two possible choices for the quotient occur. Usually, in number theory, the positive remainder is always chosen, but programming languages choose depending on the language and the signs of a and/or n.[7] Standard Pascal and ALGOL 68 give a positive remainder (or 0) even for negative divisors, and some programming languages, such as C90, leave it to the implementation when either of n or a is negative. See the table for details. a modulo 0 is undefined in most systems, although some do define it as a."

Also, in theory number the remainder take always the sign of the divisor, i don't know if it's convention or if it's for respect the number theory's laws, in particulary this law: (x+n) mod n = x mod n.

About this law, i try to demonstrate her, here's what I did:
x mod n = r with x = q*n + r où r < n et x, q, n, r are relative numbers
(x+n) mod n = r' with (x+n) = q' * n + r' r' < n et x, q', n, r' are relative numbers

So i seek to demonstrate r - r' = 0
r - r' = x - q*n - ( ( x + n ) - q'*n ) = x - q*n - x - n + q'*n = -q*n - n + q'*n = n * (-q + q' -1)

So must q = q' -1
And I do not find how to prove it. Thank you for helping me.

Also, annexment, I started trying to solve this problem with the floor.
x mod n = x - n * E(x/n)
(x+n) mod n = (x+n) - n * E((x+n)/n) = x + n - E(x/n) - 1
I've extract n/n of E((x+n)/n), it seems impossible, can you confirm ? Maybe i can't extract.

>>8266516
et = and
où = where

Sorry for this.

>>8258951

in general terms, what is a radial basis function network (what is the input and output)?

>>8266516
In math, "mod" is not a function that gives an integer. It gives what is known as "an integer mod n". The particular representative you choose for "1 mod 3" can be 1, -2, 4, -5, 7, etc. - any number whose remainder when divided by 3 is 1. If [math]x = mn + q[/math] where q is nonnegative, then [math] x + n = (m+1)n + q[/math], so they have the same remainder.

>>8266648
*(and also q < n)

>>8265791
Great

>>8265825
1 and 2 work, 3 isn't right, see below. Also, try to write proofs for everything next time

>>8266430
Impeccable

easy:
for n positive integer show that
[math] (2n+1) * \binom{2n}{n} > 4^n [/math]

medium:
n, m are positive integers less than 100.
Then show that [math] | n - m \sqrt{2} | > 1/300 [/math].

>>8266648
However, "In mathematics the result of the modulo operation is the remainder of the Euclidean division." https://en.wikipedia.org/wiki/Modulo_operation

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>>8259395
>>8259374
just because you're not supposed to use trig, doesn't mean you can't
answer (not solution) drawn in geogebra

What if I have a question from a text I am studying independently but could very well be homework for someone taking a class elsewhere using that text? Would that go in this thread?

>>8268199
I suppose you would ask about a concept pertaining to the question and given that understanding you could be better equipped to handle answering said question.

>>8268199
The point of this this thread is to not have us do your homework, but discuss interesting mathematics. If you're self studying, you fit in the second category, so go ahead and post your question.

>>8268199

Depends how you ask it.
The "I have no idea where to start" questions are the worse.
Best chance is to ask it in this thread

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>>8258960
>>8268138
blue line is parallel to (it should be obvious which line)
pink line is parallel to (it should be obvious which line)
blue angles are figured first
pink angles are figured second
should be enough info to get the red angles

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has anyone seen this yet? is it just hollywood garbage like a beautiful mind or do i get some math substance?

What does linear algebra mostly consist of?

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>>8269814
1) solving systems of linear equations
2) vector spaces

>>8269813
just think, to a supremely high genius like neumann or ramanujan, everything you consider math substance is just hollywood garbage babby tier math

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>>8269866
can we make 'doing math on the ground with chalk' the new 'doing math on windows with markers'?

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here's a fun problem if anyone's bored

very easy to write some code to check, but the mathematical explanation comes out of solution sets of certain elementary equations

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Sup /sci/, i was wondering if anyone could help me with these 2 problems:

How many ways you can sit 5 people around a table, if two of them should never be together?

and

How many ways can seven friends sit around a circular table if 3 of them should always be together?

thanks in advance

>>8271368
Is the first table circular? And are the positions of the seats important?

>>8258951
I'm taking a calculus class and as an extra
credit homework problem we were asked to integrate sqrt{1+(1/(2x)-3)^2}

Would anyone help me with this?

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>>8271999
Wow I got trips
First attempt at latex let's hope it works

so basically
$\int sqrt[1+(frac{1}{2x}-3)^2] dx$

>>8271999
>No homework

Wasted trips

>>8272010
I don't really understand what you meant to say, but I can't find the answer on mathematica and it's worth enough extra credit that it's definitely worth doing.

>>8271999
Substitute u for (2x-3)
simplify to int u/sqrt(u^2+1)
integrate to sqrt (u^2 + 1) + c
substitute (2x-3) for u
Get 4x^2 - 12x + 10 + c

Hope I didn't mess anything up

Is there such a thing as "sub-addition"? I mean an operation "below" addition in the hyperoperation chain.

I've considered one candidate already, but without any success. If we use the symbol |, we may define
a+b=a|a|...|a
with b a's. The obvious consequence is that a+1=a, but even if one makes an exception for that, other oddities crop up; one may prove that there is no identity (not even a left or right identity).

>>8272770
Nobody knows what your | symbol is supposed to mean. If you're going to make up terms, you need to tell us what you're talking about. Also, have you tried googling hyperoperation sequence? This thing is already defined, and your question amounts to "can someone read the definition for me?"

>>8269866
Actually math has progressed pretty far since them, they would probably find some modern math to be pretty difficult. And Ramanujan had sort of a poor grasp of formal methods already.

Can one of you bois recommend me a good pdf on Stochastic Calculus?

>>8272006
[eqn]\int \sqrt{1+ \left( \frac{1}{2x}-3 \right)^2} \; \mathrm{d}x [/eqn]

>>8273026
Oksendal

>>8273460
Substitute for the thing inside the ( )^2, then you essentially are finding the area of part of a circle.

https://wikimedia.org/api/rest_v1/media/math/render/svg/bf005686483ae833f1429b15cf6ecc1155851b94
Why it's not + h where h = b/2a ? It would be a little more simple isn't it ?

>>8273687
Wait...that's not right. There's a plus sign and not a minus sign. Anyways, doing a trig substitution should do the trick.

I know the gradient of a function is a vector and divergence of a vector is a scalar but are the methods of finding them really that different? Is one just specifying a particular vector and one a scalar function?

>>8259763
Yes. You can also approach from the topos-theoretic side.

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K is a field with multiplication * :K x K -> K; a is and element of K and a!=0.
Is the projection/mapping
fa: K->K, fa(b) = b*a^-1 surjective?

The stuff with the field K is confusing me and I have no idea what to do.
Anything helps

>>8276502
[math]K[/math] is a field with multiplication [math]\cdot \colon K \times K \to K[/math]; [math] a \in K, a \neq 0 [/math]
Is
[math]f_a \colon K \to K, f_a(b) = b \cdot a^{-1}[/math] surjective?

Let [math]x \in K[/math] be an arbitrary element. Then we have that [math]x \cdot a[/math] is mapped to [math]x[/math] as [math] f_a(x\cdot a) = x\cdot a \cdot a^{-1} = x\cdot (a \cdot a^{-1}) = x.[/math]

More specifically, if you want to use the definition of surjective to show that something is surjective, then what you do this is. You get an arbitrary element of the codomain. Then you show a formula to find the element of the domain that maps to that arbitrary element that you chose.

>>8273950
> I know the gradient of a function is a vector and divergence of a vector is a scalar
The gradient of a scalar field (R^n->R) is a vector (R^n), and the divergence of a vector field (R^n->R^n) is a scalar.

> but are the methods of finding them really that different? Is one just specifying a particular vector and one a scalar function?
They're both just differentiation.

Gradient takes a scalar function (x1,x2,...)->y and returns a vector of partial derivatives <dy/dx1,dy/dx2,....>. Divergence takes a vector function (x1,x2,...)->(y1,y2,...) and returns the sum of partial derivatives of same components dy1/dx1+dy2/dx2+....

Both of these are defined for any number of dimensions (unlike curl, which only exists in three dimensions).

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>>8258951
>finished high school without even doing pre-calc
>only majors I'm interested in are math-heavy
How fucked am I?

>>8276657
Learn precalc ? Takes like a week or two tops

>>8276657
>>finished high school without even doing pre-calc
>How fucked am I?
If there are highschools that graduate people without requiring precalc, we're all pretty fucked.

>>8276664
I'm just as upset as you. I'm probably getting fucked up the ass come uni.

I got a question on weak formulation.
As an example take the cauchy problem in an abstract banachspace setting:

Find [math] u \in V [/math], s.t.
[eqn] u_t = Au [/eqn]
For some operator [math] A [/math] and banach space [math] V [/math].
The weak formulation would be:
Find [math] u \in V [/math], s.t.
[eqn] (u_t,v) = a(u,v) ~~~~~~ \forall v\in V [/eqn]
where [math] a [/math] is the bilinearform induced by [math] A [/math].
My question now is, if it makes sense to choose the testspace other than [math] V [/math].
I was just reading about some finite element schemes, where you have different approximations for the test- and solution space in the sense of
Find [math] u \in U_h [/math], s.t.
[eqn] (u_t,v) = a(u,v) ~~~~~~ \forall v\in V_h [/eqn]
with [math] dim(U_h)=dim(V_h) < \infty [/math] and [math] \lim_{h\to 0}U_h=\lim_{h\to 0}V_h =V [/math].
Does a formulation like that only make sense, because both approximated spaces converge to the same space,
or is there a whole theory for nonequal test- and solutionspaces